Total enthalpy change for freezing of 1 mol of water at $10^{\circ} \mathrm{C}$ to ice at $-10^{\circ} \mathrm{C}$ is \_\_\_\_ (Given : $\Delta_{\text {fus }} \mathrm{H}=\mathrm{xkJ} / \mathrm{mol}$ $\mathrm{C}_{\mathrm{p}}\left[\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\right]=\mathrm{y} \mathrm{J} \mathrm{mol}^{-1} \mathrm{~K}^{-1}$ $\mathrm{C}_{\mathrm{p}}\left[\mathrm{H}_{2} \mathrm{O}(\mathrm{s})\right]=\mathrm{z} \mathrm{J} \mathrm{mol}^{-1} \mathrm{~K}^{-1}$

(C) $10(100 \mathrm{x}+\mathrm{y}+\mathrm{z})$

JEE Main · 2025 · 07 Apr 2025 · Shift 1 · Chemical Thermodynamics

Total enthalpy change for freezing of 1 mol of water at

1 0^{\circ} C

to ice at

- 1 0^{\circ} C

is ____

(Given :

Δ_{fus} H = xkJ / mol

C_{p} [H_{2} O (l)] = yJ mol^{- 1} K^{- 1}

C_{p} [H_{2} O (s)] = zJ mol^{- 1} K^{- 1}

Answer: (B)

Freezing proceeds in three steps. Cooling 1 mol of liquid water from

1 0^{\circ}

C to

0^{\circ}

C releases

y (0 - 10) = - 10 y J

. Freezing at

0^{\circ}

C releases

- x kJ = - 1000 x J

. Cooling the ice from

0^{\circ}

C to

- 1 0^{\circ}

C releases

z (- 10 - 0) = - 10 z J

. Adding these, the total enthalpy change is

- 1000 x - 10 y - 10 z = - 10 (100 x + y + z) J

JEE Main · 2025 · 07 Apr 2025 · Shift 1 · Single correct option